Exams › NEET › Physics

A body is executing S.H.M. When the displacements from the mean position are 4cm and 5cm, the corresponding velocities of the body are 10 cm per sec and 8 cm per sec. Then the time period of the body is

1. 2π sec
2. π√2 sec
3. π sec
4. (3π/2) sec

Correct answer: π√2 sec

Solution

In SHM, the velocity is given by v = ω√(A² - x²), where ω is the angular frequency, A is the amplitude, and x is the displacement. Using the given data for two cases, we can form two equations and solve for ω and A. From ω, the time period T = 2π/ω is calculated to be π√2 sec.

Related NEET Physics questions

• One oscillation completed by a vibrating body in one second is known as:
• A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of time t. Which of the following statements is true. Assume that the particle started at \( t=0 \) at its mean position
• The amplitude of a particle executing SHM about O is 10 cm. Then: This question has multiple correct options
• The displacement (in \( \mathrm{m} \) ) of a particle of mass 100 gram from its equilibrium position is given by \( \boldsymbol{y}=\mathbf{0 . 0 1} \sin 2 \pi(t+ \) 0.4). Select the correct option(s): This question has multiple correct options
• Three simple harmonic motions in the same direction having the same amplitude \( A \) and same period are superposed. If each differs in phase from the next by \( 45^{0}, \) then
• A particle executes \( \boldsymbol{S} \boldsymbol{H} \boldsymbol{M} \) of amplitude \( 25 c m \) and time period \( 3 s \) What is the minimum time required for the particle to move between two points \( 12.5 \mathrm{cm} \) on either side of the mean position?

⚔️ Practice NEET Physics free + battle 1v1 →